Q. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561$\ \mathring{\mathrm{A}}$. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

A) 1215 $\mathring{\mathrm{A}}$
B) 1640$\ \mathring{\mathrm{A}}$
C) 2430$\ \mathring{\mathrm{A}}$
D) 4687$\ \mathring{\mathrm{A}}$
Explanation

(a) For hydrogen or hydrogen type atoms
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$= ${RZ}^{2}\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$
In the transition from $ni \rightarrow \ nf$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \lambda\ \propto \ \frac{1}{Z^{2}\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{2}}{\lambda_{1}} = \ \frac{Z_{1}^{2}\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)_{1}}{Z_{2}^{2}\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda_{2} = \ \frac{\lambda_{1}Z_{1}^{2}\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)_{1}}{Z_{2}^{2}\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)_{2}}$
Substituting the values, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{(6561)(1)^{2}\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)}{(2)^{2}\left( \frac{1}{2^{2}}\ - \ \frac{1}{4^{2}} \right)}$ $= \ 1215\ \mathring{\mathrm{A}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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